Question

In lecture and the text, the establishment of a reference standard was made for the calculation of enthalpies for different processes. Consider the following reference standard: Foe example, the Delta H_gluc (298 K)' s for benzaldehyde (C_7H_6O), and ethanol (C_2H_6O) are: C(graphite) + C_6H_12O_6(s) rightarrow C_7H_6O(l) + 5/2 O_2(g) + 3 H_2(g) Delta H_gluc (298 K) = 1363 kJ/mol 1/3 C_6H_12O_6(S) + H_2(g) rightarrow C_2 H_6O(l) + 1/2 O_2 (g) Following the definition above and the examples provided, write the equation representing the Heat from Glucose Formation for benzoic acid, C_7H_6O_2. Calculate the Delta H_gluc(298 K) for benzoic acid and ethanol.

Answer 1

a) the equation is

C(graphite) + C6H1206 (s)    -----> C7H602 (l) + 3H2 (g) + 202 (g)

b)

now

we know that

dHgluc = dHf products - dHf reactants

so

for benzoic acid

C(graphite) + C6H1206 (s)    -----> C7H602 (l) + 3H2 (g) + 202 (g)

dHgluc = dHf C7H602 + ( 3 x dHf H2) + (2 x dHf 02) - dHf C - dHf C6H1206

dHgluc = -385.1 + ( 3 x 0) + ( 2 x 0) - 0 - ( -1273.3)

dHgluc = 888.2

so

dHgluc for benzoic acid is 888.2 kJ/mol

now

for ethanol

(1/3) C6H1206 + H2 ---> C2H60 + 0.5 02

dHgluc = dHf C2H60 + (0.5 x dHf 02) - dHf H2 - ( 1/3 x dHf C6H1206)

dHgluc = -277.69 + ( 0.5 x 0) - 0 - ( 1/3 x -1273.3)

dHgluc = 146.74

so

dHgluc for ethanol is 146.74 kJ/mol